A001448 -id:A001448 - cp's OEIS frontend

A061548 Numerator of probability that there is no error when average of n numbers is computed, assuming errors of +1, -1 are possible and they each occur with p = 1/4.

1, 3, 35, 231, 6435, 46189, 676039, 5014575, 300540195, 2268783825, 34461632205, 263012370465, 8061900920775, 61989816618513, 956086325095055, 7391536347803839, 916312070471295267, 7113260368810144185, 110628135069209194801, 861577581086657669325, 26876802183334044115405

Offset: 0

Leah Schmelzer (leah2002(AT)mit.edu), May 16 2001

			For n=1, the binomial(2*n-1/2, -1/2) yields the term 3/8. The numerator of this term is 3, which is the second term of the sequence.

Indranil Ghosh, Table of n, a(n) for n = 0..500
Robert M. Kozelka, Grade Point Averages and the Central Limit Theorem, American Mathematical Monthly. Nov. 1979 (86:9) pp. 773-7.

Cf. A001316, A001448, A061549, A117973.

Bisection of A001790.

A061548:= func< n | Numerator(Binomial(4*n,2*n)/4^n) >;
[A061548(n): n in [0..25]]; // G. C. Greubel, Oct 19 2024

seq(numer(binomial(2*n-1/2, -1/2)), n=0..20);

Table[Numerator[(4*n) !/(2^(4*n)*(2*n) !^2) ], {n, 0, 20}] (* Indranil Ghosh, Mar 11 2017 *)
Table[Numerator[SeriesCoefficient[Series[(Sqrt[1 + Sqrt[1 - x]]/Sqrt[2 - 2* x]), {x, 0, n}], n]], {n, 0, 20}] (* Karol A. Penson, Apr 16 2018 *)

for(n=0, 20, print1(numerator((4*n)!/(2^(4*n)*(2*n)!^2)),", ")) \\ Indranil Ghosh, Mar 11 2017

import math
f = math.factorial
def A061548(n): return f(4*n) // math.gcd(f(4*n), (2**(4*n)*f(2*n)**2)) # Indranil Ghosh, Mar 11 2017

def A061548(n): return binomial(4*n,2*n)/2^sum(n.digits(2))
[A061548(n) for n in (0..20)]  # Peter Luschny, Mar 23 2014

a(n) = numerator(binomial(2*n-1/2, -1/2)).
From Johannes W. Meijer, Jul 06 2009: (Start)
a(n) = numerator((4*n)!/(2^(4*n)*(2*n)!^2)).
a(n) = 2*A001448(n)/ A117973(n). (End)
a(n) = A001448(n)/A001316(n). - Peter Luschny, Mar 23 2014
a(n) is the numerator of the coefficient of power series in x around x=0 of sqrt(1 + sqrt(1 - x))/(sqrt(2)*sqrt(1 - x)). - Karol A. Penson, Apr 16 2018

More terms from Asher Auel, May 20 2001

A352373 a(n) = [x^n] ( 1/((1 - x)^2*(1 - x^2)) )^n for n >= 1.

Original entry on oeis.org

2, 12, 74, 484, 3252, 22260, 154352, 1080612, 7621526, 54071512, 385454940, 2758690636, 19810063392, 142662737376, 1029931873824, 7451492628260, 54013574117106, 392188079586468, 2851934621212598, 20766924805302984, 151403389181347160, 1105047483656041080

Offset: 1

Peter Bala, Mar 14 2022

Suppose n identical objects are distributed in 3*n labeled baskets, 2*n colored white and n colored black. White baskets can contain any number of objects (or be empty), while black baskets must contain an even number of objects (or be empty). a(n) is the number of distinct possible distributions.
Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2*n) + 2*y_1 + 2*y_2 + ... + 2*y_n.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Calculation suggests that, in fact, stronger congruences may hold.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k.
More generally, let r and s be integers and define a sequence (a(r,s;n))n>=1 by a(r,s;n) = [x^n] ( (1 + x)^r * (1 - x)^s )^n.
Conjecture: for each r and s the above supercongruences hold for the sequence (a(r,s;n))n>=1.
The present sequence is the case r = -1 and s = -3. Other cases include A000984 (r = 2, s = 0), A001700 with offset 1 (r = 0, s = -1), A002003 (r = 1, s = -1), A091527 (r = 3, s = -1), A119259 (r = 2, s = -1), A156894 (r = 1, s = -2), A165817 (r = 0, s = -2), A234839 (r = 1, s = 2), A348410 (r = -1, s = -2) and A351857 (r = -2, s = -4).

			n = 2: 12 distributions of 2 identical objects in 4 white and 2 black baskets
             White         Black
   1)   (0) (0) (0) (0)   [2] [0]
   2)   (0) (0) (0) (0)   [0] [2]
   3)   (2) (0) (0) (0)   [0] [0]
   4)   (0) (2) (0) (0)   [0] [0]
   5)   (0) (0) (2) (0)   [0] [0]
   6)   (0) (0) (0) (2)   [0] [0]
   7)   (1) (1) (0) (0)   [0] [0]
   8)   (1) (0) (1) (0)   [0] [0]
   9)   (1) (0) (0) (1)   [0] [0]
  10)   (0) (1) (1) (0)   [0] [0]
  11)   (0) (1) (0) (1)   [0] [0]
  12)   (0) (0) (1) (1)   [0] [0]
Examples of supercongruences:
a(7) - a(1) = 154352 - 2 = 2*(3^2)*(5^2)*(7^3) == 0 (mod 7^3);
a(2*11) - a(2) = 1105047483656041080 - 12 = (2^2)*3*(11^3)*13*101*103*2441* 209581 == 0 (mod 11^3).

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Paolo Xausa, Table of n, a(n) for n = 1..1000

Cf. A000984, A001448, A001700, A002003, A091527, A119259, A156894, A165817, A211419, A211421, A234839, A262733, A276098, A348410, A351856, A351857.

seq(add( binomial(3*n-2*k-1,n-2*k)*binomial(n+k-1,k), k = 0..floor(n/2)), n = 1..25);

nterms=25;Table[Sum[Binomial[3n-2k-1,n-2k]Binomial[n+k-1,k],{k,0,Floor[n/2]}],{n,nterms}] (* Paolo Xausa, Apr 10 2022 *)

a(n) = Sum_{k = 0..floor(n/2)} binomial(3*n-2*k-1,n-2*k)*binomial(n+k-1,k).
a(n) = Sum_{k = 0..n} (-1)^k*binomial(4*n-k-1,n-k)*binomial(n+k-1,k).
a(n) = binomial(4*n-1,n)*hypergeom([n, -n], [1-4*n], -1).
48*n*(n-1)*(3*n-1)*(3*n-2)*(93*n^3-434*n^2+668*n-339)*a(n) = 12*(n-1)*(21762*n^6-134199*n^5+323805*n^4-386685*n^3+237728*n^2-70336*n+7680)*a(n-1) + 5*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(93*n^3-155*n^2+79*n-12)*a(n-2) with a(1) = 2 and a(2) = 12.
The o.g.f. A(x) = 2*x + 12*x^2 + 74*x^3 + ... is the diagonal of the bivariate rational function x*t/(1 - t/((1 - x)^2*(1 - x^2))) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
A(x) = x*d/dx(log(F(x))), where F(x) = (1/x)*Series_Reversion( x*(1 - x)^2*(1 - x^2) ).
a(n) ~ sqrt(4 + sqrt(6)) * (13/4 + 31*sqrt(6)/18)^n / (2*sqrt(5*Pi*n)). - Vaclav Kotesovec, Mar 15 2022

A036910 a(n) = (binomial(4n, 2n) + binomial(2*n, n)^2)/2.

Original entry on oeis.org

1, 5, 53, 662, 8885, 124130, 1778966, 25947612, 383358645, 5719519850, 85990654178, 1300866635172, 19780031677718, 302045506654052, 4629016098160220, 71163013287905912, 1096960888092571317, 16949379732631632570, 262435310495071434602, 4070954160892425897300

Offset: 0

N. J. A. Sloane

The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1972, Eq 3.68, page 30.

G. C. Greubel, Table of n, a(n) for n = 0..500
N. J. A. Sloane, "A Handbook of Integer Sequences" Fifty Years Later, arXiv:2301.03149 [math.NT], 2023, p. 6.

Cf. A000984, A001448, A157531.

[(Binomial(4*n, 2*n) + Binomial(2*n, n)^2)/2: n in [0..40]]; // G. C. Greubel, Dec 09 2021

B[n_] := Binomial[2*n, n]/2; Table[B[2*n] + 2*B[n]^2, {n, 0, 40}] (* G. C. Greubel, Dec 09 2021 *)

a(n) = (binomial(4*n,2*n)+binomial(2*n,n)^2)/2; \\ Michel Marcus, Dec 09 2021

[(binomial(4*n, 2*n) + binomial(2*n, n)^2)/2 for n in (0..40)] # G. C. Greubel, Dec 09 2021

a(n) = Sum_{k=0..n} binomial(2*n, k)^2. - Paul Barry, May 15 2003
From G. C. Greubel, Dec 09 2021: (Start)
a(n) = (A000984(2*n) + A000984(n)^2) / 2. - corrected
a(n) = (A001448(n) + A000984(n)^2) / 2. - corrected
a(n) = A157531(2*n, n)/2. - corrected Jan 12 2023
G.f.: sqrt(1 + sqrt(1 - 16*x))/(2*sqrt(2)*sqrt(1 - 16*x)) + (1/Pi)*EllipticK(16*x). [This formula uses the conventions of Mathematica; to use it with Maple set EllipticK(4*sqrt(x)).  - Peter Luschny, Sep 05 2025]
(End)
D-finite with recurrence n^2*(2*n-1)*(n-1)*a(n) -2*(n-1) *(2*n^3+109*n^2-241*n+132) *a(n-1) +12*(-112*n^4+1008*n^3-3011*n^2+3702*n-1600) *a(n-2) +16*(832*n^4-7008*n^3+20744*n^2-24228*n+7905) *a(n-3) +256*(4*n-13) *(4*n-15)*(-7+2*n)^2*a(n-4)=0. - R. J. Mathar, Jan 12 2023
a(n) = binomial(2*n, n)^2*hypergeom([1, -n, -n], [n + 1, n + 1], 1). - Peter Luschny, Sep 04 2025

A129369 Expansion of 1/sqrt(1-4x) - x/sqrt(1-4x^2).

Original entry on oeis.org

1, 1, 6, 18, 70, 246, 924, 3412, 12870, 48550, 184756, 705180, 2704156, 10399676, 40116600, 155114088, 601080390, 2333593350, 9075135300, 35345215180, 137846528820, 538257689684, 2104098963720, 8233430022168, 32247603683100

Offset: 0

Paul Barry, Apr 11 2007

Partial sums are A129368.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000108, A000984, A001448, A002458, A129368.

B:=Binomial; [B(2*n,n) - (n mod 2)*B(n-1, Floor((n-1)/2)): n in [0..60]]; // G. C. Greubel, Jan 31 2024

CoefficientList[Series[1/Sqrt[1-4x]-x/Sqrt[1-4x^2],{x,0,30}],x] (* Harvey P. Dale, Feb 02 2012 *)

[binomial(2*n,n) - (n%2)*binomial(n-1, (n-1)//2) for n in range(61)] # G. C. Greubel, Jan 31 2024

a(n) = binomial(2*n, n) - (1/2)*(1-(-1)^n)*binomial(n-1, (n-1)/2).
From G. C. Greubel, Jan 31 2024: (Start)
a(2*n) = A001448(n).
a(2*n+1) = (n+1)*( 2*A000108(2*n+1) - A000108(n) ).
a(2*n+1) = 2*A002458(n) - A000984(n).
(n+1)^3*(n+2)*(n+3)*a(n+3) = 2*(n+1)*(n+2)^3*(2*n+1)*a(n+2) + 4*(n+1)^4*(n+3)*a(n+1) - 8*n*(n+2)^3*(2*n+1)*a(n), with a(0)=a(1) = 1, a(2) = 6. (End)

A037964 a(n) = (1/2)(binomial(4n, 2n) - (-1)^nbinomial(2*n,n)).

Original entry on oeis.org

0, 4, 32, 472, 6400, 92504, 1351616, 20060016, 300533760, 4537591960, 68923172032, 1052049834576, 16123800489472, 247959271674352, 3824345280321920, 59132290859989472, 916312070170755072

Offset: 0

N. J. A. Sloane

nonn

The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1972; Formula (3.74), page 31.

G. C. Greubel, Table of n, a(n) for n = 0..825

Cf. A000108, A000984, A001448.

[(1/2)*((2*n+1)*Catalan(2*n) -(-1)^n*(n+1)*Catalan(n)): n in [0..30]]; // G. C. Greubel, Jun 20 2022

A037964 := proc(n)
    binomial(4*n,2*n)/2-(-1)^n*binomial(2*n,n)/2 ;
end proc: # R. J. Mathar, Feb 20 2015

With[{C= CatalanNumber}, Table[(1/2)*((2*n+1)*C[2*n] -(-1)^n*(n+1)*C[n]), {n, 0, 30}]] (* G. C. Greubel, Jun 20 2022 *)

[sum(binomial(2*n, 2*k+1)^2 for k in (0..n-1)) for n in (0..30)] # G. C. Greubel, Jun 20 2022

From R. J. Mathar, Feb 20 2015: (Start)
n*(2*n-1)*(n-1)*a(n) -12*(n-1)*(4*n^2-11*n+10)*a(n-1) +4*(38*n^3-333*n^2+715*n-435)*a(n-2) +48*(34*n^3-228*n^2+499*n-355)*a(n-3) +16*(4*n-15)*(2*n-7)*(4*n-13)*a(n-4) = 0.
n*(n-1)*(2*n-1)*(5*n^2-15*n+11)*a(n) -4*(n-1)*(30*n^4-120*n^3 +161*n^2-82*n+12)*a(n-1) -4*(4*n-7)*(2*n-3)*(4*n-5)*(5*n^2-5*n+1)*a(n-2) = 0. (End)
From G. C. Greubel, Jun 20 2022: (Start)
a(n) = Sum_{k=0..n-1} binomial(2*n, 2*k+1)^2.
a(n) = (1/2)*(A001448(n) - (-1)^n*A000984(n)).
a(n) = (1/2)*((2*n+1)*A000108(2*n) - (-1)^n*A000108(n)).
G.f.: (1/4)*(1/sqrt(1+4*sqrt(x)) + 1/sqrt(1-4*sqrt(x)) - 2/sqrt(1+4*x)). (End)

A275653 a(n) = binomial(4n,2n)binomial(3n,2*n).

Original entry on oeis.org

1, 18, 1050, 77616, 6370650, 554822268, 50199951984, 4664758248000, 442077195513690, 42533571002422500, 4141601026094832300, 407220411993767798400, 40363606408574136870000, 4028061310168832261158176, 404311537318239680601595200, 40785601782042745410592271616

Offset: 0

Peter Bala, Aug 04 2016

Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)*binomial(4*n + k,4*n - k)*binomial(2*k,k)* binomial(2*n - k,n) = binomial(4*n,2*n)* binomial(3*n,2*n).
We also have Sum_{k = 0..4*n} (-1)^(n+k)*binomial(4*n + k,4*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = binomial(4*n,2*n)* binomial(3*n,2*n).
Compare with the identities
Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n + k,2*n - k)* binomial(2*k,k)*binomial(2*n - k,n) = binomial(2*n,n)^2 = A002894(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(6*n + k,6*n - k)* binomial(2*k,k)*binomial(2*n - k,n) = binomial(6*n,3*n)* binomial(2*n,n) = A275655(n)
Sum_{k = 0..n} (-1)^(n+k)*binomial(8*n + k,8*n - k)* binomial(2*k,k)*binomial(2*n - k,n) = binomial(8*n,4*n)* binomial(5*n,2*n)*binomial(2*n,n)/binomial(6*n,3*n).
See also A275652, A275654 and A275655.

Cf. A001448, A002894, A005809, A275652, A275654, A275655.

seq((4*n)!*(3*n)!/(n!*(2*n)!^3), n = 0..20);

Table[Binomial[4 n, 2 n] Binomial[3 n, 2 n], {n, 0, 15}] (* Michael De Vlieger, Aug 07 2016 *)

a(n) = (4*n)!*(3*n)!/(n!*(2*n)!^3).
a(n) = A001448(n) * A005809(n).
Recurrence: a(n) = 3*(3*n - 1)*(3*n - 2)*(4*n - 1)*(4*n - 3)/(n^2*(2*n - 1)^2) * a(n-1).
a(n) = [x^n] ((1 + x)^2/(1 - x))^(2*n) * [x^n] (1 + x)^(3*n) = [x^n] G(x)^(6*n) where G(x) = 1 + 3*x + 38*x^2 + 1150*x^3 + 47099^x^4 + 2264968*x^5 + 120311611*x^6 + ... appears to have integer coefficients.
exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^6, where F(x) = 1 + 3*x + 92*x^2 + 4579*x^3 + 282605*x^4 + 19698991*x^5 + 1484923315*x^6 + ... appears to have integer coefficients.
a(n) ~ sqrt(3/2)*108^n/(2*Pi*n). - Ilya Gutkovskiy, Aug 07 2016
From Peter Bala, Mar 23 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(3*n-k-1,n-k)*binomial(4*n,k)^2.
a(n) = [x^n] (1 - x)^(2*n) * P(4*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A275652.
The supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k. (End)

A330843 Square array T(n,k) = [x^n] ((1+x)^(k+1) / (1-x)^(k-1))^n, n>=0, k>=0, read by descending antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 2, -2, 1, 4, 6, 0, 1, 6, 30, 20, 6, 1, 8, 70, 256, 70, 0, 1, 10, 126, 924, 2310, 252, -20, 1, 12, 198, 2240, 12870, 21504, 924, 0, 1, 14, 286, 4420, 41990, 184756, 204204, 3432, 70, 1, 16, 390, 7680, 104006, 811008, 2704156, 1966080, 12870, 0

Offset: 0

Seiichi Manyama, Feb 07 2020

			Square array begins:
    1,   1,     1,      1,      1,       1, ...
    0,   2,     4,      6,      8,      10, ...
   -2,   6,    30,     70,    126,     198, ...
    0,  20,   256,    924,   2240,    4420, ...
    6,  70,  2310,  12870,  41990,  104006, ...
    0, 252, 21504, 184756, 811008, 2521260, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=1..7 give A000984, A091527, A001448, A262732, A211419, A262733, A211421.

Main diagonal is A332231.

T[n_, k_] := Sum[Binomial[(k + 1)*n, j] * Binomial[k*n - j - 1, n - j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, May 05 2021 *)

T(n,k) = Sum_{j=0..n} binomial((k+1)*n,j) * binomial(k*n-j-1,n-j).

T(n,k) = 1/n! * ((k+1)*n)!/Gamma(1 + (k+1)*n/2) * Gamma(1 + (k-1)*n/2)/((k-1)*n)!.

A357508 a(n) = binomial(4n,2n) - 2binomial(4n,n).

Original entry on oeis.org

-1, -2, 14, 484, 9230, 153748, 2434964, 37748520, 580043790, 8886848740, 136151207764, 2088760285456, 32108266614164, 494648505828904, 7637081136832840, 118158193386475984, 1831647087068431374, 28444051172077725444, 442429676097305612324

Offset: 0

Peter Bala, Oct 01 2022

Sun and Wan's supercongruence stated below apparently generalizes as follows:
Let m be an integer and k a positive integer. Define u(n) = binomial((m+2)*n,(k+1)*n) - binomial(m,k)*binomial((m+2)*n,n). We conjecture that u(n) == u(1) (mod p^5) for all primes p >= 7. [added 22 Oct 2022: the conjecture is true: apply Helou and Terjanian, Section 3, Proposition 2.]
Conjecture: for r >= 2, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ) for all primes p >= 5. - Peter Bala, Oct 13 2022

C. Helou and G. Terjanian, On Wolstenholme’s theorem and its converse, J. Number Theory 128 (2008), 475-499.
Z.-W. Sun and D. Wan, On Fleck quotients, arXiv:math/0603462 [math.NT], 2006-2007.

Cf. A001448, A005810, A357509.

seq(binomial(4*n,2*n) - 2*binomial(4*n,n), n = 0..20);

a(n) = A001448(n) - 2*A005810(n).

a(p) == -2 (mod p^5) for all primes p >= 7. (Sun and Wan, Corollary 1.5.)

A024491 a(n) = (1/(4n-1))*C(4n,2n).

Original entry on oeis.org

-1, 2, 10, 84, 858, 9724, 117572, 1485800, 19389690, 259289580, 3534526380, 48932534040, 686119227300, 9723892802904, 139067101832008, 2004484433302736, 29089272078453818, 424672260824486220, 6232570989814602524, 91901608649243484728, 1360850743459951600780

Offset: 0

Clark Kimberling

			sqrt(1/2*(1+sqrt(1-x))) = 1 - 1/8*x - 5/128*x^2 - 21/1024*x^3 - ...

Cf. A000108, A001448, A024492, A245112.

[(1/(4*n-1))*Binomial(4*n,2*n) : n in [0..20]]; // Wesley Ivan Hurt, Jan 06 2024

Table[1/(4n-1) Binomial[4n,2n],{n,0,20}] (* or *) With[{c=4Sqrt[x]}, CoefficientList[ Series[(-Sqrt[1-c]-Sqrt[1+c])/2,{x,0,30}],x]] (* Harvey P. Dale, Mar 10 2013 *)

G.f.: A(x) = -sqrt((1/2)*(1+sqrt(1-16*x))).
With interpolated zeros, this has g.f. -(sqrt(1-4x)+sqrt(1+4x))/2. - Paul Barry, Dec 23 2006
D-finite with recurrence n*(2*n-1)*a(n) - 2*(4*n-3)*(4*n-5)*a(n-1) = 0. - R. J. Mathar, Nov 13 2012
a(n) = A001448(n)/(4*n-1). - R. J. Mathar, Apr 27 2020
From Peter Bala, Apr 02 2023: (Start)
O.g.f. A(x) = - sqrt(1 - 4*x*C(4*x)), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.
The series reversion of -x*A(x) is equal to x * the o.g.f. of A245112. (End)
a(n) ~ 2^(4*n-5/2) / (n^(3/2) * sqrt(Pi)). - Amiram Eldar, Sep 04 2025

More terms from Harvey P. Dale, Mar 10 2013

A099976 Bisection of A000984.

Original entry on oeis.org

2, 20, 252, 3432, 48620, 705432, 10400600, 155117520, 2333606220, 35345263800, 538257874440, 8233430727600, 126410606437752, 1946939425648112, 30067266499541040, 465428353255261088, 7219428434016265740

Offset: 0

N. J. A. Sloane, Nov 19 2004

Vincenzo Librandi, Table of n, a(n) for n = 0..100

Cf. A000108, A000265, A000984, A001448, A002458, A063079.

[Binomial(4*n+2, 2*n+1): n in [0..20]]; // Vincenzo Librandi, May 22 2011

seq(binomial(4*n+2,2*n+1),n=0..20); # Emeric Deutsch, Dec 20 2004

Array[Binomial[4*# + 2, 2*# + 1] &, 20, 0] (* Paolo Xausa, Jul 11 2024 *)

a(n) = binomial(4n+2, 2n+1). - Emeric Deutsch, Dec 20 2004
G.f.: 2*sqrt(2)/sqrt(1-16*x)/sqrt(1+sqrt(1-16*x)) = 2 + 60*x/(G(0)-30*x) where G(k)= 2*x*(4*k+3)*(4*k+5) + (2*k+3)*(k+1)- 2*x*(k+1)*(2*k+3)*(4*k+7)*(4*k+9)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jul 14 2012
G.f. A(x) satisfies A(x^2) = F'(x)/F(x), where F(x) = C(x)/C(-x) and C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. - Peter Bala, May 15 2023
From R. J. Mathar, Jul 11 2024: (Start)
D-finite with recurrence n*(2*n+1)*a(n) -2*(4*n-1)*(4*n+1)*a(n-1)=0.
a(n) = 2*A002458(n).
G.f.: 2* 2F1(3/4,5/4; 3/2 ; 16*x).
Conjecture: A000265(a(n)) = A063079(n+1), odd part of a(n). (End)
a(n) / (2*n+2) = A024492(n). - R. J. Mathar, Jul 12 2024

More terms from Emeric Deutsch, Dec 20 2004

cp's OEIS Frontend

A061548 Numerator of probability that there is no error when average of n numbers is computed, assuming errors of +1, -1 are possible and they each occur with p = 1/4.

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A352373 a(n) = [x^n] ( 1/((1 - x)^2*(1 - x^2)) )^n for n >= 1.

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A036910 a(n) = (binomial(4*n, 2*n) + binomial(2*n, n)^2)/2.

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A129369 Expansion of 1/sqrt(1-4*x) - x/sqrt(1-4*x^2).

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A037964 a(n) = (1/2)*(binomial(4*n, 2*n) - (-1)^n*binomial(2*n,n)).

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A275653 a(n) = binomial(4*n,2*n)*binomial(3*n,2*n).

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A330843 Square array T(n,k) = [x^n] ((1+x)^(k+1) / (1-x)^(k-1))^n, n>=0, k>=0, read by descending antidiagonals.

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A036910 a(n) = (binomial(4n, 2n) + binomial(2*n, n)^2)/2.

A129369 Expansion of 1/sqrt(1-4x) - x/sqrt(1-4x^2).

A037964 a(n) = (1/2)(binomial(4n, 2n) - (-1)^nbinomial(2*n,n)).

A275653 a(n) = binomial(4n,2n)binomial(3n,2*n).

A357508 a(n) = binomial(4n,2n) - 2binomial(4n,n).